Unable To Jump To Row
I have a tariff which is atm 44 rows with 15 columns of numbers. Now i am trying to pull as individual price from the tariff based on user input: Code:
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Unable To Jump Row 0
I am getting the following error: Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 2 in /home/content/i/n/n/**/html/tracker/connect.php on line 35 Here is my coding from line 34 on down. Code:
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Pagination :: Unable To Jump To Row 0
I am running a script to find the "Previous" and "Next" records, the problem is when I get to record 1 I get the error below, same with when I reach the last record in the db. PHP Code: Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 18 Here is the script running it, do I need to add to the query? or how can I use IF/ELSE?PHP Code: <?php $id = $_GET['id']; // Retrieve your record according to id // Display Record // Links to Previous Records $query = "SELECT id FROM jokes WHERE category = '$category' AND id < $id ORDER BY id DESC LIMIT 1" ; // For Previous $result = mysql_query($query) ; ..........
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Warning: Mysql_result() [function.mysql-result]: Unable To Jump To Row 0
i want to grab the value of a sql entry. but the thing is, sometimes this value is <NULL>. i thought, no biggy, i'll just have an if statement: if (is_numeric(mysql_result($query, 0))) { //do something } else { //do something else using mysql_result($category_id_query, 0); } however, this doe not work. i get this error: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 is there a way to check to see if mysql_result() is <NULL> without throwing an error.
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Using Jump Menu
I am using a jump menu to select multiple drop down boxes in a form which are related to the choice you make in the first drop down box. The problem is that the page reloads and everything before this tag is cleared. I want to be able to repost the data. I know that I need to use sessions in my code but I dont know exactly how to use them for this case. Code:
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Jump Frame?
I want to make a jump frame (i think is the name), when people click in an image or video, appears a frame in the top (or left), but i want to put an id for each video and image. There are a simple and better way, but i don´t know how to do, where we can put only the link and it redirects with the frame in php (out.php) like this one (this is an adult link, not spam, only is an example what i want): Code:
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JUMP Out Of The Script
"mysql_connect establishes a connection for the duration of the script that access the db. Once the script has finished executing it closes the connection. The only time you need to close the connection manually is if you jump out of the script for any reason. " http://www.webmasterworld.com/forum88/903.htm What does he mean by jump out? I used header redirection, is this "jump out", but after header redirection, I always added exit;, does it mean that I need to close the connection.
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Jump Menu
<select name="menu1" onChange="MM_jumpMenu('parent',this,0)"> <option value="www.site1.com">google</option> <option value="www.site2.com">metu</option> <option value="www.site3.com">mynet</option> <option value="www.site4.com">asa</option> </select> above code ; i want to get the selected value and post as named $url into a php code. How can i call the selected value?
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Site Jump During Submission
1)I'am collecting form data on my server from a HTML form. The next steps Iam trying to figure out is: 2) I want to then go to another site (payment site) 3) then back to my site. 4) extract original data gathered from step 1 I know the simple method would be to go to pay site first then to my site. However I would like my form first! I also do not want to write the data into mysql in step 1. Not until step 4.
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Header('Location:') Jump Ignored?
I'm seeing a problem that has me flummoxed. The only thing I can think of is that I'm violating some rule I don't know about. I have some code that does some processing and then does a header('Location: ...) jump to page A on success or falls through to the jump to page B. This is the code: if ( mysql_query( 'LOCK TABLES tableX WRITE', $link ) ) { mysql_query( $q, $link ) ; // store the record $ID = mysql_insert_id( $link ) ; // save the new id if ( mysql_affected_rows($link) == 1 ) { unlock_tables() ; $_SESSION['ErrMsg'] = 'New ID is ' . $ID ; header( 'Location: PageA.php' ) ; $_SESSION['ErrMsg'] .= ' Error: ignored jump' ; } else // the store failed { error handling } } // second chance (for debugging) to do the right thing header( 'Location: PageA.php' ) ; $_SESSION['ErrMsg'] .= ' and it ignored it AGAIN!' ; // and we shouldn't get here at all -- but we do! header( 'Location: PageB.php ) ; Unfortunately, as I know from the telltales I stuff into session, the store works but the interpreter ignores both header calls that would jump to A and finally jumps to B instead. Unless I'm more tired than I'm aware, or more ignorant, this doesn't make sense.
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Jump To Section Of Same Page
PHP Group, How can I in PHP, based on a result, move the cursor to a certain location on the page? I want the same result as Anchor tags in HTML. I want to go to a certain section of the same page based on a return result in PHP. What this is for is an order forms page - when a user fills out a section incorrectly, I want the page to display at that location in the form after the user presses a submit button.
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Jump Out Of A Multidimensional Array?
trying to get out of an array dynamically. $transport = array('foot', 'bike', 'car', array('test1', 'test2')); i'm in test2( $transport[3][1] ), how do I go into $transport[3]? Is there a function capable of doing this?
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Jump Menu Selected Value
I have a set of 2 menus where the values of the 2nd are dependent on the value selected in the first. Yes, I will (eventually) convert this to a javascript process, but for now I am content reloading the page. I have created a jumpmenu that reloads the page with the state passed in the url (ie. community.php?state=NY). This determines the values displayed in a second (dependent) menu which then lists the counties available for that state. Code:
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Jump To Html Page
I am a php newbe and I am trying to figure out how to simply display an existing html page from my php script. Here is the code I use but I get the normal "output started" errors. $sql = "INSERT INTO author SET firstname='$name1', lastname='$name2', workphone='$workphone', email='$email', loginid='$loginid', password='$password'"; if (@mysql_query($sql)) { header("Location: formSubmit.htm"); <--------HERE
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Dynamic Jump Menu
I'm looking to create a dynamic jump menu/list menu that will enable me to navigate to a particular record by selecting a value from the list, however the list is dynamically generated from the recordset. Code:
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Header - Jump Menu Does Nothing In IE
I have a php header and php footer that are being called into my content pages. In the php header, I have a jump to menu system. The jump to menu works great in Firefox, however, the jump menu does nothing in IE. Have any of you experienced problems with jump to menues in php headers and whatnot?
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PHP/MySql Jump Script
I have also read in a affiliate forum (quote below)that putting &afsrc=1 at the end of a link will help click through, how might this also be implemented? "Recently, I put the &afsrc=1 on all my links across my largest sites. Damned if sales have not increased by 10-20% since then. (it has been about 2 weeks at about 13000 clicks a day)" <?php $link = mysql_connect ("localhost", "user", "pass"); mysql_select_db ("datebase"); $result = mysql_query ("SELECT link FROM id WHERE number=$locn"); $row = mysql_fetch_array($result); header("Location: " . $row["link"]); mysql_close ($link); ?>
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Jump Script And Wordpress
I created a very simple jump script, and then edited a page on my blog so the links use the jump script links. I followed a template, and triple checked all the code, so I am sure that part is right. My question is, where do I put the actual .php file so the links in the jump script actually pick up the code and go where they are supposed to go? I tried a variety of folders (public_html, themes folder, actual theme folder, etc.), but none made the script work. unless I am (quite possibly) missing something else?
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Jump To Letter In Paginated Results
I'm on a project that has paginated results coming from MySQL, say ten records on a page. Let's say that the results contain a song title. What would be the best method for me to write a script that could take a user to whichever page has the first song title starting with B? Basically, the client wants a list of letters at the bottom that take the user to the page that records beginning with the selected letter begin on. Advice? Am I going to have to loop through the results and count the positions of the letters, and then calculate which page that position would be on?
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Jump Menu Call Page
Is it possible to use a jump menu to call a page within a page? To be clearer. I can place an " include ('whatever.php'); " within my page which is fine. But i would like this include to be a variable set by the options within a jump menu. So when I select from the menu the include will swap from its default to the newly selected one. Am I making sense?
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How To Create A Jump-menu Using Dreamweaver?
I have 50 names and links I need to show up alphabetically in a jump-menu. I'm pretty green with php and databases. Is there a quick and easy way to accomplish this? I know how to create a jump-menu using Dreamweaver, but how would I set it to read the text file.
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Script That Is A Bit Like A Jump Menu But Without The Pull Down List.
I am looking for a script that is a bit like a jump menu but without the pull down list. I want users to be able to enter a post code (4 digits) and then be redirected to a web page based on that post code as each web page is different per code. the jump menu provided in dreamweaver is almost there but as there are thousands of codes it is impractical to have them drop down.
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How I Can Make A List That Acts Like A Html Jump Menu?
how I can make a list that acts like a html jump menu (for a regular list I have to use a "Go" button in order to achieve the desired behaviour and jump menus I cannot use because they work only with links to other pages). For example,I want such a list filled with questions from my db and when I select one I want to see its possible answers; the page stays the same, only its data differs based on the selection. Or you can have a look at such a jump menu on this website, the one with options for geting to various sections of the forum.
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Unable To Set Cookie!
I'm unable to set cookie! I have tried a lot of codes, but it doesn't work! I don't get any error message, but the script doesn't set cookie. Below is my code:
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Unable To Connect
Here is the setup: Server: -SQL 2000 SP3 on a win2k box -Firewall set up to allow connections for all ports ntwdblib.dll version 2.8.0 (newest) Client: XP Pro SP2 Apache 2.0.52 PHP 4.3.8.8 ntwdblib.dll version 2.8.0 (copied from server) extension=php_mssql.dll uncommented i am unable to connect to MSSQL, PHP4 through Windows XP.
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Unable To Install PHP
I compile the latest source for php, but when I try to load php module in Apache, I get this error message: Cannot load /usr/local/apache2/modules/libphp5.so into server: /usr/local/apache2/modules/libphp5.so: undefined symbol: _efree
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Unable To Insert
Might someone be able to tell me what is wrong with this query? Can't get it to send the data to the db. <? $date = date("Y-m-d G:i:s") ; $db="dbname"; mysql_query ("INSERT INTO dbname (var1, var2, var3, IP) VALUES ($_POST['var1'],$_POST['var2'],$_POST['var3'],$_POST['IP'])$db"); ?>
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Unable To Add To Database Through PHP
I have written a short file called config.php that contains all of the connectivity scripting and also a db_close.php that contains the closing script. These 2 files are used thoughout the pages that I have written, so I know that the connectivity is working fine. Here is the issue. I run a simple mysql query that inserts some values into a table. After submiting the values, the script continues as if the values have been added properly, but in fact they haven't been added at all. My problem is that I do not know why it isn't being added because it return NO error in this process. Am I supposed to add some code that returns the mysql errors? Below is the line that adds the field. If this is supposed to be posted in the MySQL forum, please move it. If more info is needed pls respond. include("scripts/config.php"); mysql_query("insert into clients values ('$clientid', '$assoc', '$fn', '$ln', '$gender', '$dob', '$phoneh', '$phonec', '$phonew', '$email1', '$email2', '$email3', '$email4', '$email5', '$fnd', '$lnd', '$fnm', '$lnm', '$phonecd', '$phonecm', '$phonewd', '$phonewm', '$fax', '$address', '$city', '$province', '$pc', '$school', '$occupation', '$balance', '$notes', '$lastlog'"); include("scripts/db_close.php");
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Unable To Upload
I'm not being able to upload at all, I tried to paste the sample included in the documentation and here is what came out: Code:
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Unable To Upload PHP Extensions
I've come across a wierd thing in my PHP configuration. Just yesterday, I upgraded everything I had. Apache to 1.3.22, PHP to 4.1, and MySQL to 3.23.46. Everything went well and I was up and running in no time and even got Zend Optimizer to work. The problem submerged when I started adding PHP extensions. The PHP 4.1 distribution contains a lot of extensions with it. I chose modules like zlib, gd, etc. For some reason I could not get LDAP module or PDF module to work. When Apache loads, I get this message:
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Unable To Connect To Certain SQL Servers From PHP On IIS.
I am unable to connect to certain SQL servers from PHP on IIS. I have two IIS 5 servers, both are Windows 2000 Server boxes with PHP 4.3.4 and MDAC 2.80.1022.3. I have three SQL Server machines - two SQL2K SP3a machines and one SQL 7.0 SP4 machine. One of the SQL2K servers is the same machine as one of the IIS 5 servers. One of the IIS servers, Server1, can mssql_connect to all three SQL servers. The other IIS server, Server2, can only mssql_connect to one of the SQL2K servers. I get a "connect failed" error when attempting to hit the other two SQL servers. The one it can connect to, incidentally, is the SQL server that resides on the _other_ IIS server. I have tried copying the PHP application from Server1 to Server2. The same PHP app that can connect to all three SQL Servers from Server1 can only connect to SQL on Server1 from Server2. Here's a chart of the servers: ServerName |OS |SQL Version |PHP Can connect to... -------------|-----------|----------------|------------------------- Server1 |Win2k Svr |SQL2K SP3a |Server1, Server3, Server4 Server2 |Win2k Svr |N/A |Server1 Server3 |NT 4.0 |SQL7 SP4 |N/A Server4 |Win2k3Svr |SQL2K SP3a |N/A
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Exec(): Unable To Fork
I am trying to use exec() on windows IIS ,but i get the following error Code: Warning: exec(): Unable to fork [html2pdf.exe "http://www.yahoo.com/" "yahoo.pdf"] in c:inetpubwwwrootoctsuppublicationpublicationreport2.php on line 62 and the code that i used is Code: <? $convertpdfprg="html2pdf.exe "http://www.yahoo.com/" "yahoo.pdf""; $checkvar=exec($convertpdfprg);?>
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Unable To Use $_SERVER['HTTP_REFERER']
Any ideas why I'm not able to use $_SERVER['HTTP_REFERER'] on my shared hosting account (PHP 4.1.2), but I can on my local server (PHP 4.3.3)? I imagine it has something to do with the different versions of PHP or perhaps php.ini, but I'm not sure what?
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Unable To Load Library?
I restarted my virtual server on Friday night and since then I've been getting these emails every 10 minutes (there's a Cronjob scheduled to execute). Problem is, I have no idea what it means or why the error is suddenly occuring.. PHP Warning: Unknown(): Unable to load dynamic library '/usr/lib/php4/pgsql.so' - libpq.so.3: cannot open shared object file: No such file or directory in Unknown on line 0
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Unable To Create A Database
I am unable to create a table in SQL using PHP. I think it might jsut be a code, because it won't even show the page. It just gives me an internal error. <?php $connection = @mysql_connect("localhost","jcruz","abc2000") or die(mysql_error()); $new_db = "testDB2"; $result = @mysql_create_db($new_db, $connection) or die(mysql_error()); ?>
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Unable To Call The Session Up
I have developed this problem as to where i am unable to call the session up in a page. The script below is the check_login.php which creates the session and all its variables. The way i am calling it is by Code:
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Pagination Unable To Navigate
i am trying to use a pagination script that is from a tutorial on here i was given a link to and basically i can get information to display on first page and it brings up standard text saying under PREV5 1 NEXT 5 but it isnt a link or interactive so i cant get next 5 records or last so i have 5 records on screen but cant navigate between them. But i cant see why the screen shows only 5 records there are no error codes can anyone see what i have done wrong please as i feel like im going round in a circle. Code:
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Unable To Delete A Record Using Php
I am having a form which displays a couple of values from database and a delete button which is used to delete that particular record. Below is the code that I am using, but I am unable to delete the record. Please help me to get the record deleted. Code:
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Unable To Display Graphics
PHP 5.1.6 installed on Windows 2000 server running IIS. In the php.ini file I've got extension_dir = "C:PHPext" and extension=php_gd2.dll I've also got C:PHP;C:PHPext set in the path environment variable. When I type in the command php -m, it lists out gd as one of the loaded modules. I now reboot the server. Browsing to a php page to display graphics (jpg's), it shows a 'no GD' icon where the graphics should be.
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Unable To Send Mail
I have a small form with 3 text fields to collect the data and send an email. Scenario 1: I am collecting the data from 3 fields and I am not using these 3 values. I am sending a HTML message which is good Scenario 2: I am collecting the data from 3 fields and appending these values in the HTML message. I am unable to send a mail from here. In both the scenarios I was using header information also. Code:
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Unable To Load Dll Mapscript_48.dll
I get this error.... Warning: dl() [function.dl]: Unable to load dynamic library '/ms4w/Apache/php/ext/php_mapscript_48.dll' - The specified procedure could not be found. in C:ms4wappsgmaphtdocsgmap75.phtml on line 56 Fatal error: Call to undefined function ms_GetVersion() in C:ms4wappsgmaphtdocsgmap75.inc.php on line 51 I'm tryin to get an application I didn't design 2 work. php_mapscript48.dll exists at the location. What do I try next? I've never dealt with dll's in php before.
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Unable To Display Errors
I am having trouble to make PHP show errors. The hosting provider of my website won't let me have access to the php.ini file and the following code also doesn't work for me: Code: ini_set('display_errors', 1); error_reporting(E_ALL); How can I force PHP to show errors anyway. The helpdesk tells me I have to be able to take care of it im my scripts.
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Unable To Find Package
I am trying to use MDB2 and can get SELECT queries to work fine, but when trying to INSERT, I am getting: MDB2 Error: not found (/usr/lib/php/PEAR.php at line 557) unable to find package 'MDB2_Datatype' file 'MDB2/Datatype.php' errors.
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Unable To Start MySQL
I have just installed, PHP, Apache, on my WinXP Pro machine. I then went to install MySQL, which appeared to go well. I went into MySQL folder and opened up winmysqladmin, and clicked on Start. the response I got was unable to start try again in 30 seconds. Don;t know if this has anything to do with it, but in the books and help pages I have got, they all mention a log on page, to enter username and password, I have not seen this.
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Unable To Connect To MySQL
I've finished coding php, run it OK with my localhost and uploaded to my Apache server with Linux OS. However, I'm seemingly unable to connect to mysql at the server side... and received the following message: Could not connect to database: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (111) What's this message abt? I tried to get into MySQL without php on the server.. and received the same message at the console... how come?
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Unable To Make URL Work?
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Unable To Create Socket
I have PHP5.1.2 & Apache on my PC and I'm trying to connect PHP via a socket to a java app acting as a server. I'm failing at the first PHP hurdle. The following line returns fatal error : Call to undefined function socket_create() in <filename>.php $socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP); Configuration: Php.ini lines default_socket_timeout = 60 extension=php_sockets.dll extension_dir = "./" ; tried with "./ext" same result... so I coppied php_sockets.dll to php5 root folder enable_dl = On ;maybe not relevant phpinfo() The only socket related items I can find are Registered Stream Socket Transports | tcp, udp default_socket_timeout | 60
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